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Oct 6, 2011 at 15:58 comment added Michael Joyce On second thought, proving that your variety is irreducible is most easily done by showing its an orbit variety, so I retract my previous approach. Instead, fix an $n+1$-diml v.s. $V$ and let $W_1, W_2$ be subspaces of dimension $m+1$ and $n-m+1$ that intersect in a $1$-diml subspace. You can find a basis, call it $\mathscr{V} = \{v_0, \dots v_n\}$ as in the above comment so that $W_1$ and $W_2$ have the form of the $H$ and $H'$ above. Then change of basis from $\mathscr{E}$ to $\mathscr{V}$ gives you an element of $PGL_{n+1}$ that takes one pair of such subspaces to the other.
Oct 6, 2011 at 15:55 vote accept IMeasy
Oct 6, 2011 at 15:53 answer added Anton Fonarev timeline score: 4
Oct 6, 2011 at 15:47 comment added Michael Joyce Fix your favorite two such subspaces $H$ and $H'$. (Mine are the span of $\{ e_0, e_1, \dots, e_m \}$ and the span of $\{ e_m, e_{m+1}, \dots, e_n \}$ for some choice of ordered basis of your underlying vector space $V$). You want to show that the $PGL_{n+1}$-orbit of this point is the variety of pairs of subspaces that intersect in a point. You know that the orbit is an irreducible subvariety of this irreducible variety, so all you need is for them to have the same dimension. You can compute the dimension of an orbit once you identify the stabilizer of a point.
Oct 6, 2011 at 14:45 history asked IMeasy CC BY-SA 3.0